Question

the sim of first n trems of an arithmetic progression is (5n-nsquare) find its nth term.

Answer 1

The sum of first n trems of an arithmetic progression is 5n-nsquare.

first term = 5*1-1^2
                =5-1
                =4

sum of first 2 terms = 5*2-2^2
                                  =10-4
                                  =6

2nd term = sum of first 2 terms - first term
    
               =6-4
 
              =2
second term = 2

hence ,first and second terms are 4,2
here, a1=4  ,  a2= 2
common difference (d) = a2-a1
         
                                     = 2-4
  
                                     = -2

we know that nth term  = a+(n-1)d
here ,

a=4
d=-2
n=n

hence, nth term = a+(n-1)d
 
                          =4+(n-1)(-2)
 
                          =4-2n+2
 
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Answer 2

We should calculate the  n th term without explicitly calculating the  first term and common difference.

Sn  - S_n-1  = Tn
So   Tn = 5n - n²  - 5(n-1) + (n-1)²
So   Tn = 5 -2 n + 1 = 6 - 2 n